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A recently asked question here was solved with the claim that any symmetric square matrix $M$ of the following form is positive definite:

  • All of the off-diagonal elements are the same positive integer $k$.

  • Each diagonal element is a positive integer $n_i \gt k$. The diagonal elements may or may not be equal to one another.

The matrix arises as the product of a particular incidence matrix with its transpose. Why is a matrix of this form positive definite?

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  • $\begingroup$ Thanks, but in this case positive definite was essential to the proof. $\endgroup$ – Robert Shore Feb 26 at 0:34
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    $\begingroup$ It is positive definite because it is the sum of a positive definite matrix (namely, the positive diagonal matrix $\operatorname{diag}(n_1-k,\ n_2-k\ldots)$) and a positive semidefinite matrix (whose elements are all equal to $k$). $\endgroup$ – user1551 Feb 26 at 0:39
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    $\begingroup$ @user1551 Elegant argument. I was messing around with row-echelon form. You ought to post this as an answer, I think. $\endgroup$ – saulspatz Feb 26 at 0:47
  • $\begingroup$ @RodrigodeAzevedo We needed to know the matrix is positive definite in order to get control over the rank of the incidence matrix. $\endgroup$ – Robert Shore Feb 26 at 0:53
  • $\begingroup$ @user1551 I agree, it's answer time! Ready to upvote it b/c you beat me to it. $\endgroup$ – Oscar Lanzi Feb 26 at 1:25
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It is positive definite because it is the sum of a positive definite matrix (namely, the positive diagonal matrix $\operatorname{diag}(n_1−k,\ n_2-k,\ldots)$) and a positive semidefinite matrix (whose elements are all equal to $k$).

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